Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12
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A kac moody verma module is not integrable in general

Statement refuted

False claim: a Kac–Moody Verma module of dominant integral highest weight is integrable.

For every finite GCM with at least one simple index and every λ, its Verma highest vector satisfies fiNvλ0 for every N0 and every simple index i. Thus no such Verma module is integrable, even at dominant integral λ. The concrete witness is A=[2], λ=0.

Facts & Assumptions

Given: A finite GCM with at least one simple index and its Verma module.

[F1]

Verma modules are nonzero highest-weight modules with highest vector vλ (Universal property and pbw character of kac moody verma modules).

[F2]

Integrability requires local nilpotence of every fi (Integrable kac moody module).

[F3]

The induced tensor construction identifies MA(λ) with U(n), sending vλ to 1 (Kac moody verma module).

[F4]

Ordered monomials form a basis for the enveloping algebra, with homogeneous bases available without AC (PBW for countably presented Kac Moody Lie algebras).

[F5]

The simple negative root space is the nonzero line Cfi (Kac moody root spaces are finite dimensional).

Counterexample

1.1

Fix a simple index i. By F5 its simple negative root space is one dimensional with specified nonzero vector fi. Use the homogeneous basis in F4, replacing its vector at that degree by fi if needed; this is a single nonzero rescaling. For every N0, the ordered monomial consisting of N copies of fi is a PBW basis monomial by F4, including the empty monomial 1 at N=0. Hence these powers are nonzero and independent in U(n). F3 identifies them with fiNvλ, so none vanishes.

F3F4F5given
2.1

By 1.1, the single nonzero vector vλ from F1 is killed by no power of fi. This fails the local nilpotence requirement F2 and proves nonintegrability. For the explicit witness A=[2], λ=0, the label zero is dominant integral, while v0,fv0,f2v0, are all nonzero independent vectors. Thus the extra dominant-weight hypothesis does not repair the false claim. This includes the zeroth and first powers; there is no terminal power. A GCM with no simple indices is excluded, since then there is no lowering operator to witness failure. The basis construction and one specified rescaling require no AC.

F1F2step 1.1

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